Solving a Parallel-Plate Capacitor Problem

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In summary, the electric field in the capacitor is 0.614 cm^2 and the speed of the electron is 5.20 x 10^6 m/s.f
  • #1

Homework Statement

The figure below shows an electron entering a parallel-plate capacitor with a speed of v = 5.20 x 10^6 m/s. The electric field of the capacitor has deflected the electron downward by a distance of d = 0.614 cm at the point where the electron exits the capacitor. The length of the capacitor is 2.25 cm. (a) Find the magnitude of the electric field in the capacitor. (b) Find the speed of the electron when it exits the capacitor.

Homework Equations

I tried to solve for the electric field using E=kq/r^2 but my answer was off by orders of magnitude. For the radius, I used the length of the capacitor/2 b/c the electron ends up making a parabolic shape. For the final velocity, I tried using the equation v^2=Vo^2+2gdelta y but my answer was not correct for that either.
  • #2
What is the force exerted by the electric field E on a charge q? Since the E field is fairly uniform in between the plates of a parallel plate capacitor, what does that say about the force?

The E field equation that you used is not applicable in between capacitor plates.
  • #3
If the field of E is fairly uniform between the plates of a parallel plate capacitor, does that mean that E=0 ?
  • #4
No, there is a relationship for the differential charge on the two plates "Q", the capacitance "C", and the voltage across the capacitor "V". Do you know what that relationship is? And remember that the units of electric field are V/m, so if you know the voltage across a capacitor, you would know the electric field between the plates, right?

But in this problem, you calculate the electric field E inside the parallel plate capacitor based on the effects it creates on the electron, not from some capacitor geometry information (which you are not given anyway). All you need to keep in mind is that the E field is uniform (sometimes called a "flat" field) in the volume between the plates of the capacitor.

What kinematic motion equations do you think you can use, given that the E is uniform? What was that equation that I was asking for that involves F, q and E?
  • #5
So the relationship between Q, C, and V is C=Q/V, correct? and would the relationship between F, q, and E be F=Eq ? but if that is the case, then I would have to calculate F and what would I use for the two q values?
  • #6
You have the correct equations, but remember that I said you don't calculate the E field from the first equation -- you use kinematics and the second equation to calculate E.

You have a constant E acting on the charge q while it moves through the capacitor. What does F=qE tell you about the force's direction and magnitude in terms of E? If that force is constant (is it?), then you can use the simple kinematic equations of motion for the charge q. Write the equations for the horizontal and vertical motion of the electron as a function of the time it spends flying through the capacitor, and solve for the questions in the problem.

I have to bail now for a few hours. Those kinematic equations are the key.

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